Problem: Let $f(x)=\dfrac{\sin(x^2)}{3x}$. Find $f'(x)$. Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{2\cos(x^2)}{3}$ (Choice B) B $\dfrac{2x^2\cos(x^2)-\sin(x^2)}{3x^2}$ (Choice C) C $\dfrac{2x^2\sin(x^2)-\cos(x^2)}{3x^2}$ (Choice D) D $\dfrac{2x\cos(x^2)}{3}$
Answer: $f$ is a quotient of a composite function and another function. Let... $u(x)=\sin(x)$ $v(x)=x^2$ $w(x)=3x$... then $f(x)=\dfrac{u\Bigl(v(x)\Bigr)}{w(x)}$. To find $f'(x)$, we will need to use the quotient rule and the chain rule! $\begin{aligned} &\phantom{=}f'(x) \\\\ &=\dfrac{d}{dx}\left[\dfrac{u\Bigl(v(x)\Bigr)}{w(x)}\right] \\\\ &=\dfrac{\dfrac{d}{dx}\left[u\Bigl(v(x)\Bigr)\right]w(x)-u\Bigl(v(x)\Bigr)w'(x)}{[w(x)]^2} \gray{\text{Quotient rule}} \\\\ &=\dfrac{u'\Bigl(v(x)\Bigr)v'(x)w(x)-u\Bigl(v(x)\Bigr)w'(x)}{[w(x)]^2} \gray{\text{Chain rule}} \end{aligned}$ Let's differentiate $u$, $v$, and $w$ : $u'(x)=\cos(x)$ $v'(x)=2x$ $w'(x)=3$ Now we can plug the equations for $u$, $v$, $w$, $u'$, $v'$, AND $w'$ into the expression we got: $\begin{aligned} &\phantom{=}\dfrac{{u'\Bigl(v(x)\Bigr)}v'(x)w(x)-{u\Bigl(v(x)\Bigr)}w'(x)}{[w(x)]^2} \\\\ &=\dfrac{{\cos(x^2)}(2x)(3x)-{\sin(x^2)}(3)}{(3x)^2} \\\\ &=\dfrac{6x^2\cos(x^2)-3\sin(x^2)}{9x^2} \\\\ &=\dfrac{2x^2\cos(x^2)-\sin(x^2)}{3x^2} \gray{\text{Simplify}} \end{aligned}$ In conclusion, $f'(x)=\dfrac{2x^2\cos(x^2)-\sin(x^2)}{3x^2}$.